# Uniform integrability and Lebesgue convergence

A). Given that $|X_n| \leq Y$ and $Y \in L$. Try to show $X_n$ is lebesgue integrable.

b). Try to give any example for which $X_n \longrightarrow^{L} X$ yet $\not\exists Y \in L$ with $|X_n| \leq Y$.

c). Try to give any example for which $X_n \to X$ w.p.1, and $X_n, X \in L$ yet $X_n \not\longrightarrow^L X$.

d). If $X_n$ is uniform integrable, does it follow that $g(X_n)$ is uniform integrable if g is continuous?

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Maybe $L$ denote the set of integrable random variables defined on a probability space. In b), you have to give more precisions about what $X_n\to X$ means (in which sense), even if we can guess it. – Davide Giraudo Oct 28 '12 at 10:00
Dear David, I have rewrite my description. In part b) what I mean is $X_n \longrightarrow^L X$. – abrocod Oct 28 '12 at 17:48

1. There is no need of the parameter $n$: if $0\leq X\leq Y$ and $Y$ is integrable, it follows from the definition of Lebesgue integral that $X$ is integrable. If $0\leq s\leq X$ is a simple function, then $0\leq s\leq Y$ so $\sup\{\int S,0\leq S\leq X,S\mbox{ simple}\}$ is finite.

2. Try $X_n:=\sqrt n\chi_{((n+1)^{—1},n^{-1})}$.

3. Try $X_n:=n\chi_{((n+1)^{—1},n^{-1})}$.

4. Take $f$ an integrable function which is not in $L^2$. Then $X_n(x):=f(x+n)$ and $g(x)=x^2$.

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Thanks for your answer, David. I have two wonderings: Is $\chi$ in part 2 means chi-square density? And how to show g(x) is not uniform integrable in part 4? – abrocod Oct 28 '12 at 22:50
No, it's the characteristic function of a set. You have to show that $\{f(\cdot +n)^2\}$ is not uniformly integrable. If it was, it would be bounded $L^1$, so each function would be in $L^1$. – Davide Giraudo Oct 28 '12 at 22:52
hi, david, can you please be more specific about what does $\chi_{(n+1)^{-1}, n^{-1}}$ mean? – abrocod Oct 29 '12 at 1:20
Sorry, parenthesis where missing. $\chi_A(x)$ is $1$ if $x\in A$ and $0$ otherwise. – Davide Giraudo Oct 29 '12 at 9:15